Mechanical Properties of Solids

Here, for steel wire,
Area of cross-section, A1=3.0×10−5m2
Stretching, Δl1=Δl (say)
Stretching force on steel, F1=F
For copper wire, length of wire, l2=3.5m
Area of cross-section, A2=4.0×10−5m2
Stretching, Δl2=Δl (given);
Stretching force on copper, F2=F
Let Y1 and Y2 be the Young's modulus of steel and copper wire respectively.
and Y2=

F2×A2

Δl2∕l2

=

F2×l2

A2×Δl2

=

F×3.5

4×105×Δl

...(ii)
Dividing (i) by (ii), we get
=

18.8

10.5

=1.79=1.8
⇒Y1:Y2=1.8:1